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Cumulative Distribution function

Im trying to find the cumulative distribution function F(x)=P(X<=x) of X, where X is continuous and its p.d.f is defined by

f(x)=x/6+1/4(x^2) (0<x<2) and 0 otherwise

I know you integrate it but what do you put as you limits?

Reply 1

zeratul

generally

F(x)=\int\limits_{-\infty}^x f(t)dt

in this case you discuss:
if x<=0, then F(x)=0
if 0<x<2, then

F(x)=\int\limits_{0}^x f(t)dt

if 2<=x, F(x)=1.

However, if you don't get that

\lim\limits_{x\rightarrow 2} F(x)=1

then you have a big problem because your pdf won't integrate to 1 over the whole R.
Also could you be more specific next time about your second term in f(x)?
the way I see it, that means

\frac{x^2}4

, but then why write it 1/4(x^2)?
if you meant

\frac{1}{4x^2}

, then you should be more careful

(edited 12 years ago)

Reply 2

Slim123

ok i did mean (x^2)/4.

so what is the answer then is it 1? because i thought that was always the case (integral of a pdf for x values is always 1, thats what defines a pdf)

Reply 3

zeratul

in that case i am very sorry
well, one could assume there was a part a in the question where you were asked to find c such that x/6+cx^2 is a valid pdf for 0<x<2 and 0 ow.